In order to make \(S\) more than level-\(K\), there are some groups of positions, each of which should contain the same alphabet. By counting the number of occurrence of each alphabet in each group, and choosing the alphabet which occurs most frequently as the alphabet for the group, one can accomplish the minimum number of necessary changes (in most cases).

However, it is possible that the level-0 part of a level-\(K\) part becomes a palindrome. In such case, the outcome of such modifications do not satisfy the condition of begin exactly level-\(K\), so one must handle such cases by choosing one of the level-0 part (except for the one letter in the center) and choose the second most frequent alphabet in the group.

In the implementation, it repeats the operation of folding \(S\) into half for \(K\) times, then perform calculation for the remaining level-0 part.
The total time complexity is \(\Theta(|S|)\).

Sample Code (C++)

#include <iostream>
#include <string>
#include <numeric>
#include <vector>
#include <valarray>
using namespace std;
void chmin(int& a, int b){ if(a > b) a = b; }

int main(){
    int K;
    string S;
    cin >> K >> S;
    int N = S.size();
    vector c(N, valarray(0, 26));
    for(int i = 0; i < N; i++) c[i][S[i] - 'a']++;

    if(K > 20 || (N << 1 >> K) == 0 || (N >> K) == 1) return puts("impossible") & 0;
    
    int ans = 0, x = 0;
    for(; x < K; x++){
        const int L = N / 2, odd = N % 2;
        for(int i = 0; i < L; i++){
            c[i] += c.back();
            c.pop_back();
        }
        if(odd){
            ans += (1 << x) - c.back().max();
            c.pop_back();
        }
        N = L;
    }
    if(N == 0){
        cout << ans << endl;
        return 0;
    }
    {
        const int L = N / 2, odd = N % 2;
        int pal = 0x3fffffff;  // The cost for avoiding palindrome
        for(int i = 0; i < L; i++){
            auto& a = c[i];
            auto& b = c.back();
            vector<int> ia(26), ib(26);
            iota(ia.begin(), ia.end(), 0);
            iota(ib.begin(), ib.end(), 0);
            partial_sort(ia.begin(), ia.begin() + 2, ia.end(), [&](int x, int y){ return a[x] > a[y]; });
            partial_sort(ib.begin(), ib.begin() + 2, ib.end(), [&](int x, int y){ return b[x] > b[y]; });
            if(ia[0] != ib[0]) pal = 0;
            ans += (2 << x) - a[ia[0]] - b[ib[0]];
            chmin(pal, a[ia[0]] - a[ia[1]]);
            chmin(pal, b[ib[0]] - b[ib[1]]);
            c.pop_back();
        }
        if(odd) ans += (1 << x) - c.back().max();
        ans += pal;
    }
    cout << ans << endl;
}

Sample Code (Python)

from collections import Counter

K = int(input())
S = input()

N = len(S)
if N << 1 >> K == 0 or N >> K == 1:
    exit(print("impossible"))

S = [Counter({c : 1}) for c in S]
ans = 0
for x in range(K):
    L = N // 2
    odd = N % 2
    for i in range(L):
        S[i] += S[-1]
        S.pop()
    if odd:
        ans += (1 << x) - S[-1].most_common(1)[0][1]
        S.pop()
    N = L

if N == 0:
    exit(print(ans))

x = K
L = N // 2
odd = N % 2
pal = 1 << 30  # The cost for avoiding palindrome
for i in range(L):
    a = S[i].most_common(2)
    b = S[-1].most_common(2)
    S.pop()
    if a[0][0] != b[0][0]:
        pal = 0
    a = [a[0][1], a[1][1]] if len(a) == 2 else [a[0][1], 0]
    b = [b[0][1], b[1][1]] if len(b) == 2 else [b[0][1], 0]
    ans += (2 << x) - a[0] - b[0]
    if pal > a[0] - a[1]:
        pal = a[0] - a[1]
    if pal > b[0] - b[1]:
        pal = b[0] - b[1]
if odd:
    ans += (1 << x) - S[-1].most_common(1)[0][1]
ans += pal
print(ans)